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📅 May 14, 2026 👤 scraface

May 14, 2026

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📌 We seek the sum of all positive divisors $ d \mid 1440 $ such that $ d \equiv 1 \pmod{4} $.

📌 A divisor $ d $ satisfies $ d \equiv 1 \pmod{4} $ if it is odd (so $ d $ not divisible by 2) and $ d \equiv 1 \pmod{4} $. So restrict to odd divisors. Since $ d $ must be odd, we ignore the power of 2. So consider only divisors of $ 3^2 \cdot 5 = 45 $.

📌 The odd divisors of 1440 are exactly the divisors of 45. List them:

📌 1, 3, 5, 9, 15, 45

📌 Now reduce each modulo 4:

📌 $ 1 \equiv 1 \pmod{4} $

📌 $ 3 \equiv 3 \pmod{4} $

📌 $ 5 \equiv 1 \pmod{4} $

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